Question: Find the limit as $x$ approaches negative infinity. $\lim_{x\to-\infty}\dfrac{5x^3-3x}{\sqrt{4x^6-7}}=$
Explanation: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the numerator is $x^3$, let's divide by $x^3$. In the denominator, let's divide by $-\sqrt{x^6}$, since for negative values, $x^3=-\sqrt{x^6}$. $\begin{aligned} &\phantom{=}\lim_{x\to-\infty}\dfrac{5x^3-3x}{\sqrt{4x^6-7}} \\\\ &=\lim_{x\to-\infty}\dfrac{\dfrac{5x^3-3x}{x^3}}{\dfrac{\sqrt{4x^6-7}}{-\sqrt{x^6}}} \gray{\text{Divide sides by }x^3=-\sqrt{x^6}} \\\\ &=\lim_{x\to-\infty}-\dfrac{\dfrac{5x^3-3x}{x^3}}{\dfrac{\sqrt{4x^6-7}}{\sqrt{x^6}}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to-\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to-\infty}-\dfrac{\dfrac{5\cancel{x^3}}{\cancel{x^3}}-\dfrac{3\cancel x}{\cancel x \cdot x^2}}{\sqrt{\dfrac{4\cancel{x^6}}{\cancel{x^6}}-\dfrac{7}{x^6}}} \\\\ &=\lim_{x\to-\infty}-\dfrac{5-\dfrac{3}{x^2}}{\sqrt{4-\dfrac{7}{x^6}}} \\\\ &=\lim_{x\to-\infty}-\dfrac{5-0}{\sqrt{4-0}} \gray{\lim_{x\to-\infty}\dfrac{k}{x^n}=0} \\\\ &=-\dfrac{5}{\sqrt{4}} \\\\ &=-\dfrac{5}{2} \end{aligned}$ In conclusion, $\lim_{x\to-\infty}\dfrac{5x^3-3x}{\sqrt{4x^6-7}}=-\dfrac{5}{2}$.